Meeting w3 chapter 2 part 1
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Transcript of Meeting w3 chapter 2 part 1
Chapter 2 Analog Control SystemEddy Irwan Shah Bin SaadonDept. of Electrical EngineeringPPD, [email protected]
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Outline:1. Introduction2. Laplace Transform – Table/ Theorem/ Eg.3. Common Time Domain Input Function4. Transfer Function – Open/ Closed Loop & Eg.5. Electrical Elements Modelling – Table & Eg.6. Mechanical Elements Modelling - Table & Eg.7. Block Diagram Reduction - Table & Eg.8. System Response – Poles/ Zeros, Second
Order, Steady State Error, Stability Analysis
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1. Intro - Objective of this chapter
After completing this chapter you will be able to:
Describe the fundamental of Laplace transforms.
Apply the Laplace transform to solve linear ordinary differential equations.
Apply Mathematical model, called a transfer function for linear time-invariant electrical, mechanical and electromechanical systems.
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2. What is Laplace Transform?
Laplace transform is a method or techniques used to transform the time (t) domain to the Laplace/frequency (s) domain
What is algebra & calculus?
Time Domain Frequency Domain
Differential equations
Input q(t)
Output h(t)
Algebraic equations
Input Q(s)
Output H(s)
Calculus Algebra
Laplace Transformation
Inverse Laplace Transformation
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Laplace Transform (cont.)
The Laplace transform solution consists of the following three steps:
(1) the Laplace transformation of q1(t) and (r dhldt + h = Gq) to frequency domain
(2) the algebraic solution for H(s)
(3) the inverse Laplace transformation of H(s) to time domain h(t).
(4) The calculus solution is shown as step 4.
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Definition of the Laplace Transform
Laplace transform is defined as
Inverse Laplace transform is defined as
)(tf )()(0
sFdtetf st
L
L-1
j
j
st tfdsesFj
sF
)()(
2
1)]([
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Laplace Theorem
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Laplace Table
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Example 1
Find the Laplace transform for 1)( tfSolution:
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Example 2
Find the Laplace transform for atetf )(
Solution:
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Example 3
Find the inverse Laplace transform of
Solution: 21
32
10)(
ssssF
Expanding F(s) by partial fraction:
Where,
Then, taking the inverse Laplace transform
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Example 4
Given the ,solve for y(t) if all initial conditions are zero. Use the Laplace transform method.
Solution:Substitute the corresponding F(s) for each term:
Solving for the response:
Where, K1= 1 when s=0 K2=-2 when s=-4 K3= 1 when s=-8
Hence
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3. Common Time Domain Input Functions Unit Step Function
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Unit Ramp Function
cont.
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Unit Impulse Function
cont.
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4. Transfer Function
Definition:Ratio of the output to the input; with all initial conditions are zero
If the transformed input signal is X(s) and the transformed output signal is Y(s), then the transfer function M(s) is define as;
From this,
Therefore the output is
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TF of Linear Time Invariant Systems In practice, the input-output relation of lines time-invariant system
with continuous-data input is often described by a differential equation
The linear time-invariant system is described by the following nth-order differential equation with constant real coefficients;
c(t) is output
r(t) is input
).()(
....)()(
)()(
.....)(
)(
)(
011
1
1
011
1
1
trbdt
tdrb
dt
trdb
dt
trdb
tcadt
tdca
dt
tcda
td
tcda
m
m
mm
m
m
n
n
nn
n
n
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cont.
Taking the Laplace transform of both sides,
If we assume that all initial conditions are zero, hence
Now, form the ratio of output transform, C(s) divided by input transform. The ratio, G(s) is called transfer function.
).(___)(....)()(
)(___)(.....)()(
01
1
01
1
trofconditioninitialsRbsRsbsRsb
tcofconditioninitialsCasCsasCsam
mm
m
nn
nn
)().....()().....( 011
1011
1 sRbsbsbsbsCasasasa mm
mm
nn
nn
)....(
)....(
)(
)()(
011
1
011
1
asasasa
bsbsbsb
sR
sCsG
nn
nn
mm
mm
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cont.
The transfer function can be represented as a block diagram
General block diagram
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Block Diagram of Open Loop System
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Block Diagram of Closed Loop System
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Example 1
Problem: Find the transfer function represented by
Solution:
Taking the Laplace transform of both sides, assuming zero initial conditions, we have
The transfer function, G(s) is
)()(2)(
trtcdt
tdc
)()(2)( sRsCssC
2
1
)(
)()(
ssR
sCsG
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Example 2
Problem: Use the result of Example 1 to find the response, c(t), to an input, r(t)=u(t), a unit step and assuming zero initial conditions.
Solution:Since r(t)=u(t), R(s)=1/s, hence
Expanding by partial fractions, we get
Finally, taking the inverse Laplace transform of each term yields
)2(
1)()()(
sssGsRsC
2
2/12/1)(
ss
sC
tetc 2
2
1
2
1)(
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Example 3 Problem: Find the transfer function, G(s)=C(s)/R(s), corresponding to the
differential equation
Solution:
rdt
dr
dt
rdc
dt
dc
dt
cd
dt
cd34573
2
2
2
2
3
3
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Example 4
Problem: Find the differential equation corresponding to the transfer function,
Solution:
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12)(
2
ss
ssG
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Example 5
Problem: Find the ramp response for a system whose transfer function is,
Solution:
)8)(4()(
ss
ssG
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